Question

Heights of men on a baseball team have a​ bell-shaped distribution with a mean of and a standard deviation of . Using the empirical​ rule, what is the approximate percentage of the men between the following​ values? a.166 cm and 202 cm b. 172cm and 196cm

Answer 1

Let assume that the mean is 184 and the standard deviation is 6

Heights of men on a baseball team have a​ bell-shaped distribution with a mean 184 of and a standard deviation of 6 . Using the empirical​ rule, what is the approximate percentage of the men between the following​ values? a.166 cm and 202 cm b. 172 cm and 196cm

Answer:

P(156<X<202) =  99.7%

P(172<X<196) =  95.5%

Step-by-step explanation:

Given that :

Heights of men on a baseball team have a​ bell-shaped distribution with a mean of and a standard deviation of . Using the empirical​ rule, what is the approximate percentage of the men between the following​ values? a.166 cm and 202 cm b. 172 cm and 196cm

For a.

Using the empirical​ rule, what is the approximate percentage of the men between the following​ values 166 cm and 202 cm.

the z score can be determined by using the formula:

$z = \dfrac{X - \mu}{\sigma}$

$z(166) = \dfrac{166-184}{6}$

$z(166) = \dfrac{-18}{6}$

z(166) = -3

$z(202) = \dfrac{202-184}{6}$

$z(202) = \dfrac{18}{6}$

z(202) = 3

P(156<X<202) = P( μ - 3σ < X < μ + 3σ )

P(156<X<202) = P( - 3 < Z < 3)

P(156<X<202) = P( Z <  3) - P(Z < -3)

P(156<X<202) = 0.99865-  0.001349

P(156<X<202) =  0.997301

P(156<X<202) =  99.7%

For b.

b. 172 cm and 196cm

$z = \dfrac{X - \mu}{\sigma}$

$z(172) = \dfrac{172-184}{6}$

$z(172) = \dfrac{-12}{6}$

z(172) = -2

$z(196) = \dfrac{196-184}{6}$

$z(196) = \dfrac{12}{6}$

z(196) = 2

P(172<X<196) = P( μ - 2σ < X < μ + 2σ )

P(172<X<196) = P( - 2 < Z < 2)

P(172<X<196) = P( Z <  2) - P(Z < -2)

P(172<X<196) = 0.9772 -  0.02275

P(172<X<196) =  0.95445

P(172<X<196) =  95.5%